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Reasoning: My guess would have to be that it has to do with the strength of the intermolecular forces between each molecule. For example, molecular compounds that have a very low molar mass like $\ce{CO}$ and $\ce{CH4}$. Both of theses molecules have a very small molar mass, a very low London dispersion force, and thus a very low boiling point. This in turns makes it easier to break the bonds that hold each individual atom and more than likely cause them to turn into a gas in normal conditions.

Then when you look at ionic compounds like $\ce{MgCl2}$ and $\ce{NaCl}$, it is more likely for these compounds to become solid because of the strong intermolecular force that they have (ionic bonds).

Is my reasoning correct and is there something else that I should consider?

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  • $\begingroup$ Are you specifically interested in crystalline solids or just solids in general? The title implies the former but there doesn't seem to be any mention of it in the question. $\endgroup$ – bon May 22 '15 at 17:22
  • $\begingroup$ @bon Reading my entry again, it would seem that you're right. I will change it to solids because is what I initially had in mind. $\endgroup$ – Luis Averhoff May 22 '15 at 17:32
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You are pretty much on the mark with looking at the strength of intermolecular forces. If you wanted to make a general comparison, the tendency to form a solid can be arranged by the strength of the intermolecular forces.$$\text{London Forces}<\text{Dipole-Dipole}<\text{Hydrogen bonding}<\text{Metallic}<\text{Ionic}<\text{Network Covalent}$$ Then, to compare within a group, you need some way of measuring the strength of that type of intermolecular force.

A more mathematical way to look at this if you have empirical values for a compound is to look at the $\Delta G_\text{fusion}$ at room temperature. $\Delta G_\text{fusion}=\Delta H_\text{fusion}-T\Delta S_\text{fusion}$, where $\Delta H_\text{fusion}$ is negative (since it will free up energy in forming the solid) and $\Delta S_\text{fusion}$ is negative (since, whether it started as a gas or a liquid, forming a solid will reduce entropy). Letting $\Delta G_\text{fusion}=0$ (reaction at equilibrium), we find that $T_{\text{fusion}}=\frac{\Delta H_\text{fusion}}{\Delta S_\text{fusion}}$. This allows us to see that when a compound becomes solid is a trade-off between the enthalpy released from forming bonds and the energy needed to reduce entropy at a given temperature.

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